A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

Given, side of the square field = 10 m
Therefore, perimeter = 10 m × 4 = 40 m
Farmer moves along the boundary in 40 s
Time = 2 minutes 20 s = 2 × 60 s + 20 s = 140 s
since, in 40 s farmer moves 40 m
Therefore, in 1s distance covered by farmer = 40 ÷ 40 = 1m.
Therefore, in 140s distance covered by farmer = 1 × 140 m = 140 m
Now, number of rotation to cover 140 along the boundary = 𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
= 140 m ÷ 40 m = 3.5 round
Thus after 3.5 round farmer will at point C (diagonally opposite to his initial position) of the field.
Therefore, Displacement AC= √(〖10〗^2+〖10〗^2 ) =√200 = √(10&2) m
Thus, after 2 minute 20 second the displacement of farmer will be equal to√(10&2 )m north east from initial position.

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